This document provides instructions and questions for a structural design exam. It consists of 4 questions. Students must answer question 1 and any other two questions. Question 1 involves calculating bending moments, designing reinforcement, and determining shear capacity for concrete beams. Question 2 involves checking the adequacy of steel sections and designing a bolt connection. Question 3 uses force methods to determine reactions and draws shear and bending moment diagrams. Question 4 analyzes a frame under vertical and lateral loads to determine reactions and internal forces at specific points. The document also includes relevant design formulas and appendices on load combinations, bending moment coefficients, and steel design strengths.
This section examines a method of analysing the bending moment and shear force distribution along a uniformly loaded, simply supported beam. The method used is Macaulay’s and the stages involved in this method are described in detail. The method also determines the position of the maximum bending moment, the slope and deflection at this point.
Poisson’s ratio is the ratio between lateral and axial strain in a directly loaded system. The derivation of standard equations and their further development is shown.
This section examines a method of analysing the bending moment and shear force distribution along a uniformly loaded, simply supported beam. The method used is Macaulay’s and the stages involved in this method are described in detail. The method also determines the position of the maximum bending moment, the slope and deflection at this point.
Poisson’s ratio is the ratio between lateral and axial strain in a directly loaded system. The derivation of standard equations and their further development is shown.
An Improved Subgrade Model for Crash Analysis of Guardrail Posts - University...Altair
In this study, a computer simulation model of a rigid impactor loading laterally a roadside W152x13.4 post has been developed. The interaction of cohesionless soil with a post was studied and compared to an existing dynamic test results from a published literature. Two approaches to simulate the soil have been studied: the continuum method where the soil is modeled as a solid element with Drucker and Prager material law, and the subgrade method where the soil reaction is simulated by a series of nonlinear springs. An improved method of the subgrade approach has been developed where the soil is modeled as a system of parallel springs and dampers with a lumped mass attached to the post. A simple procedure to calculate the lumped soil mass and the damping coefficient is presented.
An Improved Subgrade Model for Crash Analysis of Guardrail Posts - University...Altair
In this study, a computer simulation model of a rigid impactor loading laterally a roadside W152x13.4 post has been developed. The interaction of cohesionless soil with a post was studied and compared to an existing dynamic test results from a published literature. Two approaches to simulate the soil have been studied: the continuum method where the soil is modeled as a solid element with Drucker and Prager material law, and the subgrade method where the soil reaction is simulated by a series of nonlinear springs. An improved method of the subgrade approach has been developed where the soil is modeled as a system of parallel springs and dampers with a lumped mass attached to the post. A simple procedure to calculate the lumped soil mass and the damping coefficient is presented.
Prestress loss due to friction & anchorage take upAyaz Malik
This document provides a detailed procedure for calculating prestress loss due to anchorage take-up. Prestress Loss due to friction is also discussed in detail.
Spring 2015 problems for the course Rak-43.3110 Prestressed and precast concrete structures, Aalto University, Department of Civil and Structural Engineering. European standards EN 1990 and EN 1992-1-1 has been applied in the problems.
1. CITY UNIVERSITY OF HONG KONG
Course code & title : BST10346
Structural Design
Session : Semester B, 2007-2008
Time allowed : Two hours
This paper has SEVENTEEN pages (including this cover page).
1. This paper consists of four questions.
2. Answer Question 1 and ANY OTHER TWO questions.
3. Start each question on a new page.
4. State all the assumptions clearly if necessary
5. ALL STEPS OF CALCULATION AND SIGN CONVENTION MUST
BE CLEARLY SHOWN.
Materials, aids and instruments permitted to be used during examinations:
Approved Calculator
Special materials (other than standard materials e.g. answer book or
supplementary sheet) to be supplied to students:
Nil
Not to be taken away
1
2. Question 1 (40 marks)
The bending moment diagram of the continuous beams below from different load cases are
shown below:
190 200
Dead Load kNm
180 220 230
150 90
Live Load kNm
130 180 200
a) Calculate the ultimate design moment at mid-span and support. (5 marks)
b) Calculate the As required and As’ required (if any) of the mid-span (10 marks)
moment only for the two concrete section sizes with the following data
provided.
1.) 400 x 600 mm depth
2.) 400 x 500 mm depth
fcu = 45 N/mm²
fy and fyv = 460 N/mm²
Nominal cover = 30 mm
Link diameter = 12 mm
Main bar diameter = 40 mm
Shear link diameter = 12 mm
c) Suggest three methods in order to avoid provision of compression steel. (3 marks)
d) From the result of b), design the shear reinforcement for the two (5 marks)
sections size if the shear force from DL and LL is 300 kN and 200 kN
respectively.
2
3. e) Calculate the vc of the below section. (5 marks)
Design data:
fcu = 40 N/mm²
fy and fyv = 460 N/mm²
d = 900 mm
Link diameter and spacing
= T12@250c/c 2 legs
+
T16@200c/c 1 leg
f) Calculate the maximum shear capacity in kN of the beam. (10 marks)
g) If only two legs are allowed in this beam, design the other shear (2 marks)
reinforcement for this beam.
3
4. Question 2 (30 marks)
a) Prove that 254 x 254 x 73 kg/m Grade 55C UC and 305 x 165 x 54 (18 marks)
kg/m Grade 55C UB are adequate against the following ultimate load
and ultimate moments about X-X axis in low shear load case.
Mxx = 100 kNm
Myy = 30 kNm
F = 700 kN
b) From the question a), discuss why UB has sufficient structural capacity (2 marks)
against the same loadings as UC although UB has lighter self weight.
c) Design the bolt connection detail according to the diagram below to (10 marks)
fulfill the following ultimate load. The shear plane will be happened in
the threads.
(Remarks: The no. of bolts should not be less than 6)
4
5. Question 3 (30 marks)
a) Use force method to determine all reactions for the beam shown below. (15 marks)
b) Sketch the bending moment and shear force diagrams, indicating (12 marks)
maximum and minimum values, and exact location of contra-flexure
point(s), where appropriate.
M (kNm)
L (m) L (m)
c) If the roller support is removed from the beam, sketch the (3 marks)
corresponding bending moment and shear force diagrams.
(Hint: No further calculation shall be required)
5
6. Question 4 (30 marks)
A single-bay frame is subject to a vertical UDL of 5 kN/m and a horizontal point load of
20 kN as shown in the figure below. Using an approximate approach, determine, under the
combined vertical and horizontal loads:
a) the reaction forces at the hinge supports D and E (indicate clearly the (15 marks)
direction of reaction force);
b) the bending moments and shear forces at point B (middle of beam) and (15 marks)
point C (right end of beam). Indicate clearly whether the moment is
sagging or hogging, and whether the shear is clockwise or anti-
clockwise.
L = 4m
5 kN/m
20 kN
A B C
H = 4m
D E
Hint:
♦ For lateral load analysis, you can assume a point of contra-flexure at mid-span of the
beam.
♦ For vertical load analysis, you can assume a point of contra-flexure located at 0.1L from
each end of the beam of span L.
♦ Moment: Sagging Hogging
♦ Shear: Clockwise Anti-clockwise
- END -
6
7. Formulae
K = M / fcu bd²
z = (0.5 + √(0.25 - K/0.9) )d
As req. = M / (0.87fyz)
As′ = (K - 0.156) fcu bd² / ((0.87 fy (d - d′)
As = (0.156 fcu bd²) / (0.87 fy 0.775d) + As′
Mu (conc) = 0.156 fcu bd²
Mu (steel) = 0.87fyAsz
z = d - 0.45x
x = (0.87fyAs) / (0.45fcub0.9)
v = V/bvd
vc = 0.79(100As/bvd)1/3(400/d)1/4/1.25(fcu/25)1/3
Asv/sv = 0.4bv/(0.87fyv)
Asv/sv = bv(v-vc)/(0.87fyv)
Msx = αsx nlx²
Msy = αsy nly²
Fs/Ps + Ft/Pt ≤ 1.4
Ft = Pe y1 / no. of bolt column x Σy²
F/Aepy + Mx/Mcx + My/Mcy ≤ 1
Mc = S py but ≤ 1.2pyZ for plastic or compact sections
Mc = pyZ for semi-compact and slender sections
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8. Appendices
Load Ultimate Limit State
Combination Dead Imposed Wind Earth & Serviceability
water Limit State
pressure
Dead & Imposed 1.4 (or 1.6 - 1.4 1.0
(& earth & water 1.0)
pressure)
Dead & wind (& 1.4 (or - 1.4 1.4 1.0
earth & water 1.0)
pressure)
Dead & imposed 1.2 1.2 1.2 1.2 1.0
& wind (& earth
& water
pressure)
Conditions:
1. The area of each bay exceeds 30m²,
2. Spans are considered to be approximately equal if they do not differ by more than 15% of
the longest span.
3. Qk/Gk < 1.25,
4. Qk < 5 kN/m² excluding partitions
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9. ly / lx 1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0
αsx 0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118
αsy 0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029
Bending moment coefficient for slabs spanning in two directions at right-angles, simply-
supported on four sides
9